đkxđ: ....

\(\sqrt{x+4}+\sqrt{x+11}=x+27-x^2\)

\(\Leftrightarrow x+4+2\sqrt{\left(x+4\right)\left(x+11\right)}+x+1=x^2+729+x^4+54x-2x^3-54x^2\)

\(\Leftrightarrow2x+5+2\sqrt{\left(x+4\right)\left(x+11\right)}=x^4-2x^3-53x^2+54x+729\)

\(\Leftrightarrow2\sqrt{x^2+15x+44}=x^4-2x^3-53x^2+52x+724\)

\(\Leftrightarrow2\sqrt{x^2+15x+44}=\left(x-2\right)\left(x^3-53x-54\right)+616\) 

.........

\(a,\left(x^2-4x+11\right)\left(x^4-8x^2+21\right)=35\)

Phương trình trên tương đương với:

\(\left[\left(x-2\right)^2+7\right]\left[\left(x^2-4\right)^2+5\right]=35\left(1\right)\)

Do: \(\hept{\begin{cases}\left(x-2\right)^2+7\ge7\forall x\\\left(x^2-4\right)^2+5\ge5\forall x\end{cases}}\Rightarrow\left[\left(x+2\right)^2+7\right]\left[\left(x^2+4\right)^2+5\right]\ge35\forall x\)

Nên: \(\left(1\right)\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2+7=7\\\left(x^2-4\right)^2+5=5\end{cases}\Leftrightarrow}x=2\)

Vậy ..................................

\(b,\sqrt{x}+\sqrt{1-x}+\sqrt{x\left(1-x\right)}=1\)

\(Đkxđ:0\le x\le1\) Đặt: \(0< a=\sqrt{x}+\sqrt{1-x}\Rightarrow\frac{a^2-1}{2}=\sqrt{x\left(1-x\right)}\)

\(+)\) Phương trình mới là: \(a+\frac{a^2-1}{2}=1\Leftrightarrow a^2+2a-3=0\Leftrightarrow\left(a-1\right)\left(a+3\right)=0\)

\(\Leftrightarrow a=\left\{-3;1\right\}\Rightarrow a=1>0\)

\(\sqrt{x}+\sqrt{1-x}=1\)

\(+)\) Nếu \(a=1\Leftrightarrow x+1-x+2\sqrt{x\left(1-x\right)}=1\Leftrightarrow\sqrt{x\left(1-x\right)}=0\)

\(\Rightarrow x=\left\{0;1\right\}\left(tm\right)\)

Vậy .............................

Đk: \(\forall x\in R\)

Ta có:\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

<=> \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=\sqrt{1+2020^2+2.2020+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(1+2020\right)^2+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\frac{2021^2-2020}{2021}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=2021\)

Lập bảng xét dầu

x                   -2                   1 

x - 1   -         |           -          0       +

x + 2   -        0         +          |            -

Xét các TH xảy ra :

TH1: x \(\le\)-2 => pt trở thành: 1 - x - x - 2 = 2021

<=> -2x = 2022 <=> x = -1011 (tm)

TH2: \(-2< x\le1\) => pt trở thành: 1 - x + x + 2 = 2021

<=> 0x = 2018 (vô lí) => pt vô nghiệm

TH3: \(x>1\) => pt trở thành: x - 1 + x + 2 = 2021

<=> 2x = 2020 <=> x = 1010 (tm)

Vậy S = {-1011; 1010}

Lời giải:

a. ĐKXĐ: $x\geq 0$

$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$

$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$

$\Leftrightarrow 13\sqrt{2x}=28$

$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$

$\Leftrightarrow 2x=\frac{784}{169}$

$\Leftrightarrow x=\frac{392}{169}$

b. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$

PT $\Leftrightarrow \frac{3x-2}{x+1}=9$

$\Rightarrow 3x-2=9(x+1)$

$\Leftrightarrow x=\frac{-11}{6}$ (tm)

\(4x^2-5x-4\sqrt{x-1}-2=0\left(x\ge1\right)\)

\(\Leftrightarrow\left(4x^2-4x+1\right)-\left(x-1+4\sqrt{x-1}+4\right)=0\)

\(\Leftrightarrow\left(2x-1\right)^2-\left(\sqrt{x-1}+2\right)^2=0\)

\(\Leftrightarrow\left(2x-1-\sqrt{x-1}-2\right)\left(2x-1+\sqrt{x-1}+2\right)=0\)

\(\Leftrightarrow\left(2x-\sqrt{x-1}-3\right)\left(2x+\sqrt{x-1}+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=2x-3\\\sqrt{x-1}=-\left(2x+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x\in\varnothing\end{matrix}\right.\)

Vậy với x = 2 thì thỏa mãn pt